3 Answers
3

Rings that satisfy the condition
$R^n \cong R^m \iff n=m$ are said to have invariant basis number or the invariant basis property. P. M. Cohn has constructed examples of rings which fail to have this property, even giving examples of (non commutative) integral domains for which e.g. $R^3\cong R$ but $R^2\neq R$.

Suppose $R^n\cong R^m$ for some $n,m$. Then there must exist integers $h,k$ such that

A ring satisfying this condition is of type $(h,k)$. In [P. M. Cohn, Some remarks on the invariant basis property, Topology 5 (1966), 215-228] a fairly simple construction is given for rings of type $(h,k)$ for any $h,k$.

Certainly we can do algebraic K-theory over rings without the invariant basis property; we just need to be a little more careful. For example we won't necessarily have $K_0(R)\cong {\bf Z} \oplus \tilde{K}_0(R)$.

There are examples of exotic behavior like that which you propose. The specific objects which you should look for are the Leavitt algebras of type (2,2). A very good source on how to create many such examples is George Bergman's paper "Coproducts and some universal ring constructions" although there are easier methods for the specific example you seek.